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PH and pOHHere is the pH of each of the following:pH = log[H3O+] , pH + pOH = 14, [H3O+] * [OH-] = 1.0 X 10^-14pH = -log(0.001) = 3then...pOH = -log[OH-] = -log(0.001) = 3pH = 14- pOH = 14 - 3 = 1
Could you show me a few of the steps between line one and two in both of these conclusions?
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